For a Banach Space $E$, the Banach Alaoglu Bourbaki theorem asserts that the closed unit ball in $E^*$: $$B_{E^*}= \{f \in E^* \ | \ ||f|| \leq 1 \} $$
is compact in the weak$^*$ topology $\sigma(E^*,E)$. I have already seen the proof of the theorem. What about the ball that is not unitary?
The paper I am reading is using the fact that every closed ball (Of any radius) is compact, but it is never explicitly shown. Is this trivial corolary of the Banach-Alaoglu-Bourbaki theorem? Do I have to use the definition of open coverings to show it? This would be trivial if we were talking about the usual metric topology given by the norm, but dealing with the weak$^*$ topology is not that clear.
If $f_0\in X^*$, $\lambda>0$, and you define $T:X^*\to X^*$ by $$Tf=f_0+\lambda f$$then $T$ is continuous in the weak* topology.